# Using $\epsilon$-$\delta$ approach to prove that $\lim_{(x,y,z)\rightarrow (0,0,0)}\frac {y^3-1000xy^2+z^5}{x^2+y^2+z^4}=0$?

As the title indicates, Using $\epsilon$-$\delta$ approach to prove that $$\lim_{(x,y,z)\rightarrow (0,0,0)}\frac {y^3-1000xy^2+z^5}{x^2+y^2+z^4}=0$$?

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Here is a method without spherical coordinates, only crude inequalities. First, we have:

$$\left| \frac{y^3-1000xy^2+z^5}{x^2+y^2+z^4} \right| \leq \frac{|y|^3}{x^2+y^2+z^4}+1000\frac{|x|y^2}{x^2+y^2+z^4}+\frac{|z|^5}{x^2+y^2+z^4}.$$

Since each of $x^2$, $y^2$ and $z^4$ is non-negative, we have $x^2+y^2+z^4 \geq y^2$ and $x^2+y^2+z^4 \geq z^4$, so that:

$$\left| \frac{y^3-1000xy^2+z^5}{x^2+y^2+z^4} \right| \leq \frac{|y|^3}{y^2}+1000\frac{|x|y^2}{y^2}+\frac{|z|^5}{z^4} = |y|+1000|x|+|z|.$$

Hence, the limit at $(0,0,0)$ of this function is $0$.

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Let $$\begin{eqnarray} x &=& r\cos\phi\cos\theta &=& rau &\qquad& a &=& \cos\phi &\qquad& u &=& \cos\theta \\ y &=& r\cos\phi\sin\theta &=& rav &\qquad& b &=& \sin\phi &\qquad& v &=& \sin\theta \\ z &=& r\sin\phi &=& rb \end{eqnarray}$$ Then we have $$\lim_{r \to 0} \frac{r^3a^3v^2(v-1000u)+r^5b^5}{r^2a^2+r^4b^4} = \lim_{r \to 0} \frac{r a^3v^2(v-1000u)+r^3b^5}{ a^2+r^2b^4} = 0$$ which vanishes everywhere (and is thus well-defined). To see this, consider first that when $a\neq0$, the denominator is dominated by the $a^2$ term so that the limit becomes $$\lim_{r \to 0} \; rav^2(v-1000u) = 0$$ while when $a=0$ (i.e. at the north and south poles where $b=\pm1$), the limit becomes $$\lim_{r \to 0} \frac{r^3b^5}{r^2b^4} = \lim_{r \to 0} \; rb = 0$$ which is still independent of the direction of approach to the origin.

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I don't understand why you write the last equality: $\lim_{r\to 0}\frac{r^3b^5}{r^2b^4}=\lim_{r\to 0} r\sin\phi=0$... –  tetrapharmakon Jan 23 '12 at 10:46
Thank you @tetrapharmakon, i fixed my algebraic error, and the limit is $0$. –  bgins Jan 23 '12 at 12:13